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Electromagnetic Waves are coupled time-varying electric and magnetic fields that propagate in space. This varying electric field gives rise to a magnetic field, this magnetic field keeps changing with respect to time and it gives rise to an electric field and this process continues. In the figure below, the red line represents the electric field and it differs in the form of a sine wave. The magnetic field is represented by the blue lines in the figure. The magnetic field is also a sine wave but it is perpendicular to the electric field. They together form the electromagnetic field.

If the electric field moves along the x-axis and the magnetic field moves along the y-axis, then the wave will propagate in the z-axis. Magnetic field and electric field are perpendicular to each other and to the direction of propagation of waves. The electric and magnetic fields which are time-varying and coupled to each other give rise to electromagnetic waves.

## Sources of Electromagnetic Waves (EM)

Electromagnetic Waves are a combination of electric and magnetic field waves produced by moving charges. EM waves are created by the oscillation of electrically charged particles (accelerating charges). The electric field associated with the accelerating charge vibrates due to which a vibrating magnetic field is generated. These vibrating electric and magnetic fields give rise to EM waves. Both the electric and magnetic fields in an electromagnetic wave will fluctuate in time, one causing the other to change.

## Nature of Electromagnetic Waves RE

In an electromagnetic wave, the electric field and magnetic field are perpendicular to each other and at the same time are perpendicular to the direction of propagation of the wave, this nature of EM wave is known as Traverse nature. In these traverse waves, the direction of disturbance or displacement in the medium is perpendicular to that of the propagation of the wave. The particles of the medium oscillate in a direction perpendicular to the direction of propagation of the wave. Because of this EM waves are transverse in nature.

The electric field of the EM wave is represented as

EY = E0ψ (kx – ωt)

where,
EY = electric field along the y-axis,
x = direction of propagation of the wave
Wave number k = (2π / λ).

The magnetic field of the EM wave is represented as

BZ = B0ψ (kx – ωt)

where,
BZ = magnetic field along the z-axis
x = direction of propagation of the wave
Wave number k = (2π / λ)

## Displacement Current

Electricity and magnetism are related to each other. As the electric current travels through a wire, it creates magnetic field lines around the wire. This type of current is called conduction current, which is created by the movement of electrons through a conductor such as an electrical wire. Whereas a displacement current is a type of electrical current needed to make electromagnetic waves, it is important for the propagation of electromagnetic waves. The displacement current produces in space due to a change of electric flux linked with the surface.

This law is stated as when an electromotive force is induced in a coil or a circuit when there is a magnetic field varying with time, or there is a rate of change in magnetic flux through the coil. Using Lenz’s law we can determine the direction of the induced emf. It suggests that the induced emf always opposes the cause of its formation in the first place.

The relation between the emf ε in a wire and the electric field E in the wire is given by,

ε = ∫ E . dl

Where, dl is the element of the contour of the surface, combining this with the definition of flux,

φB = ∫ B . dA

The integral form of the above equation can be written as,

∫ E . dl = -d/dt ∫ B . dA

## Maxwell-Ampere Law

The progress in the theory of displacement current can be traced back to a famous physicist named James Clerk Maxwell. Maxwell is well known for Maxwell’s equations. The combination of four equations demonstrates the fundamentals of electricity and magnetism. For displacement current, we will be focusing on one of these equations known as the Maxwell-Ampere law.

Before Maxwell, Andre-Marie Ampere had developed the famous equation known as Ampere’s law. This law relates the magnetic field (B) surrounding a closed loop to the conduction current (I) traveling through that loop multiplied by a constant known as the permeability of free space (μ0).

∫B . ds = μ0I

Whenever there is continuous conduction current Ampere’s law holds true, but there are cases when problems arise in the law as it’s written. For example, a circuit with a capacitor in it. When the capacitor is charging and discharging, current flows through the wires creating a magnetic field, but between the plates of the capacitor, there is no presence of current flow. According to Ampere’s law, there can be no magnetic field created by the current here, but we know that a magnetic field does exist. Maxwell realized this discrepancy in Ampere’s law and modified it in order to resolve the issue.

∫B . ds = μ0 (I + ε0 (dφE /dt))

This final form of the equation is known as the Maxwell-Ampere law.

The part Maxwell added to it is known as displacement current (Id), and the formula is,

Id = ε0 (dφE /dt)

The above equation consists of two terms multiplied together. The first is known as the permittivity of free space (ε0), and the second is the derivative with respect to time and electric flux (φE). Electric flux is the rate of flow of an electric field through a given area. By taking its derivative with respect to time, we consider the change in that rate of flow over time.

## Displacement Current using Maxwell’s Equation

Maxwell adjusted the main relation of Ampere’s Circuital Law with an additional term. This made the relationship complete and wholesome with both static and time-varying parts present to play their part. Determination of displacement can be done using,

At first, take the magnetic field intensity without the magnetism,

B = μH

∇ × H = J + Jd

Performing the divergence action,

∇ . (∇ × H) = 0 = ∇ . J + ∇ . Jd

To define the Jd term,

∇ . Jd = -∇ . J

∇ . Jd = ∂pv / ∂t           …(1)

Using Gauss’s law, ∇ . D = p, Where D is the displacement vector and p is the charge density. Equation (1) becomes,

∂pv / ∂t = (∂ / ∂t) (∇ . D)

= ∇ . (∂D / ∂t)

So, ∇ . Jd = ∇ . (∂D / ∂t)

Therefore,  Jd = ∇ . (∂D / ∂t)

This derived term is known as the displacement current density formula. This comes in use when time-varying fields are required. For a constant displacement vector, i.e. a constant charge density the displacement current density vanishes.

So, the integral form of Maxwell’s equation is,

∫E . da = Q / ε0

∫B . da = 0

∫E . dl = -∫δB / δt . (da)

∫B . dl = μ0 l + μ0ε0 ∫(∂E / ∂t)

## Need for Displacement Current

Ampere’s circuital law for conduction of current during charging of a capacitor was found inconsistent. Therefore, Maxwell modified Ampere’s circuital law by introducing the concept of displacement current.

Displacement currents play a central role in the propagation of electromagnetic radiation, such as light and radio waves through empty space. It is required to make the conduction current lead in the circuit. Conduction current in wires can be made to lead the voltage by means of displacement current inside the capacitor and it has vast uses in induction motors, industrial appliances, and in our day-to-day life.

## Solved Problems on Displacement Current

Problem 1: Instantaneous displacement current of 2.0 A is set up in the space between two parallel plates of a 1 μF capacitor. The rate of change in potential difference across the capacitor is.

Solution:

In a capacitor of capacitance C,

V = q/C

dV/dt = i/C = 2.0A / 1μF = 2 x 106 V/s

Problem 2: A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current I. Consider a plane surface of area A/4 parallel to the plates and drawn between the plates. What is the displacement current through this area?

Solution:

Electric field between the plates is given as: E= q/Aε0  = It / Aε0

So the electric flux through the area A/4 is given by,

φE = (A/4)E =  It / 4ε0

Then the displacement current will be,

ID = ε0 (dφE / dt) =  ε0 d/dt (It / 4ε0) = I/4

Hence, the displacement current through this area is I/4.

Problem 3: A parallel plate capacitor with circular plates of radius (R) is being charged. At the instant, the displacement current in the region between the plates enclosed between R/2 and R is given by

Solution:

Displacement current is given by

I = ε0 (dφE / dt)

= ε0 A(dE / dt)    (Since,  φ=A.E)

= A (d/dt) (q/A)

= (A/A) i           (Since,  dq/dt = i)

= π(R/2)2 / πR2

I   = 1/4 i

Problem 4: A coil that has 700 turns develops an average induced voltage of 50 V. What must be the change in the magnetic flux that occur to produce such a voltage if the time interval for this change is 0.7 seconds?

Solution:

Given:

Number of turbine (n) = 700

Induced voltage (e) = 50 V

Time interval (dt) = 0.7 s

By using Faraday’s law we get,

e = N (dφ/dt)

Therefore, change in flux (dφ) is given as,

dφ = e×dt / N  =  50×0.7 / 700

dφ = 0.05 Wb

Problem 5: The magnetic flux linked with a coil having 250 turns is changed from 1.4 Wb to 2 Wb in 0.45 seconds. Calculate the induced emf in the coil.

Solution:

Given:

Number or turns (N) = 250

Initial flux (φ1) = 1.4 Wb

Final flux (φ2) = 2 Wb

Therefore, change in flux (dφ) can be given by,

dφ = φ2 – φ1  =  2 – 1.4  =  0.6 Wb

Time interval (dt) = 0.45 s

According to Faraday’s law we have,

Induced emf (e) = N (dφ/dt) = 250 × (0.6 / 0.45)

e = 333.33 V

Problem 6: A 0.20 m wide and 0.60 m long rectangular loop of wire is oriented perpendicular to a uniform magnetic field of 0.30 T. What is the magnetic flux through the loop?

Solution:

Given:

Length of rectangular loop (L) = 0.60 m

Breadth of rectangular loop (B) = 0.20 m

Area of rectangle (A) = 0.60 x 0.20 = 0.12 m2

Magnetic field (B) = 0.30 T

we know,

Magnetic flux (φ) = BA

φ = 0.3 × 0.12 = 0.036 Wb

## FAQs on Displacement Current

Question 1: State the factors affecting the propagation of electromagnetic waves.

The velocity of an electromagnetic wave is a property that is dependent on the medium in which it is traveling. Other properties such as frequency, time period, and wavelength are dependent on the source that is producing the wave.

Question 2: What are electromagnetic waves

Electromagnetic waves are also known as EM waves. Electromagnetic radiation consists of EM waves that are produced when an electric field comes in contact with the magnetic field.

Question 3: State some applications of electromagnetic waves

A few applications of electromagnetic waves are:

• Electromagnetic waves can transmit energy in a vacuum or using no medium at all.
• Electromagnetic waves play a very crucial role in communication technology.
• EM waves are used in radars.
• UV rays are a form of electromagnetic waves which are used to detect forged bank notes.
• Infrared radiation is used for night vision and is used in the security camera.

Question 4: What is displacement current? Obtain an expression of displacement current for a charged capacitor.

Displacement current is the current in the insulated region due to the changing electric flux.

For a charged capacitor, the electric field between the plates is given by:

E = Q / ε0 A

Q = ε0 φE

Displacement current is given by,

id = dQ/dt = ε0 (dφE /dt)

Question 5: Explain the Necessity of Displacement current